Not so fast here. Those ellipse are the problem. Cantor applies toan enumerated set. It can be infinte but it must be countable. If youassume the set is uncountable then you needn't worry about Cantor butif you assert it countable it falls to you to show how to count the membersand Cantor's diagonalization will apply.

> Here, the ellipse in the middle has that the values go from zero, to> > one, here with EF or BT (EF_oo and BT_oo, where BT_oo is EF_2^oo).> > > > Then, with the antidiagonal argument, starting from the beginning, and> > different at a place for each, "the" antidiagonal:> > > > .111(1) = 1.0> > > > is at the "end" of the list of those values.

There is no end and the diagonalization differs from every indexedpath at the indexed node. Since you are talking about an "infinitebinary tree" there is only one way to traverse the nodes to achievea particular path and the nth node differs from that of the diagonalbecause it's value differs from the value at the nth node.

> With Cantor's first, also called nested intervals, with that there is> > a way to naturally or lexicographically order the expansions, these> > are the same elements as all that comprise the complete ordered field,

Not so fast here either. Cantor didn't order the set of rationals innumeric order when he produced his bijection with the set of naturalnumbers and he certainly didn't just assert there was a bijection.

> because they are between zero and one, and there are none not in it.

But are the countable?

> Here, via the total ordering of the elements in the path, and that> > they are only constructed in that order, they go from zero to one, and> > then as inputs to the algorithm of nested intervals: there aren't> > any.> > > > Then of course there are the powerset results, for these there is a> > set-theoretic recourse, about how the reals aren't just the complete> > ordered field, but the continuum, then also about how the infinite is.> > > > Then, for each node in the tree, finitely distant from the root, and> > there are infinitely many: those are countable. Here, Cantor's> > results about the binary tree, are not via a direct anti-> > diagonalization, instead, that the reals biject to to {0,1}^w as does> > P(N) that transitively they have the same cardinal (Cantor-Schroeder-> > Bernstein theorem generally), and that the paths do to the Cauchy> > expansions of the unit line segment, of real numbers.> > > > |{0,1}^w| = |R_[0,1]| = |R| = |{0,1}^w| = |P(N)| > |N|> > > > It would be of interest to see an actual diagonalization of the tree,> > the constructive argument. Because, in the breadth-first traversal of> > the paths, the order of the expansions is the same as the order of the> > natural integers.

There is not breadth-first traversal of "the paths" because every breadth-first traversal is of uncountably infinite paths. Every nodehas an infinite set of paths passing through it. This is always trueno matter how big the node number. The set of paths passing througha given node is never finite.

> Draw a line: that's what Cantor shows: the line is drawn without> > lifting the pencil. It's drawn: from the origin, its origin.> > > > There are infinite integers, are there not: infinite integers?